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  2. Cubicle Stack #2

    Only 13 this time.
  3. Today
  4. Digging Probabilities

    Question 1: Questions 2 and 3 are correct! Followup question: What are the chances I get the health from the 14th dig versus the 15th dig? Do they differ significantly?
  5. Yesterday
  6. Born on a Wednesday

    In that case, a revised #2:
  7. Born on a Wednesday

    @Molly Mae ...
  8. Born on a Wednesday

    Yeah, I'm pretty sure the above can't be right.
  9. Born on a Wednesday

    I'm hoping it's something from a probability perspective. I assume an equal likelihood of being born on any day of the week. I don't know if I need to state that assumption. As I mentioned before, I'm not terribly great at this.
  10. Born on a Wednesday

    @Molly Mae
  11. Born on a Wednesday

    In that case:
  12. Cubicle Stack #2

    @Thalia
  13. Cubicle Stack #2

    CEX
  14. Cubicle Stack #2

    @Thalia
  15. Cubicle Stack #2

    @Thalia
  16. Cubicle Stack #2

    @Thalia
  17. Digging Probabilities

    My understanding of the puzzle If that's all true, then Q1: Q2: Q3:
  18. Cubicle Stack #2

    Recount. EEM and EEC intentionally left out. About CEX
  19. Jelly beans join the clean plate club

    @Molly Mae maybe this'll help. Q: How should you arrange all those plates of jelly beans? A: Put them on a table.
  20. Born on a Wednesday

    @Molly Mae ...
  21. Jelly beans join the clean plate club

    I lied. I can show you two numbers whose average is even that cannot be reduced to their average (and thus not reduced to 0) without a third plate. The example of {4,8}. These two alone cannot be reduced to 0. As soon as you add in a third plate, regardless of the number on that plate, you can reduce one of them to 0. That is the cornerstone of my reasoning, but I'm not certain I can express it.
  22. Jelly beans join the clean plate club

    I sure don't! But I also couldn't find an easy way to express the relationship between (1) the average of two numbers being even and (2) the third plate. That's when I disappeared down the rabbit hole of the difference between a and b being divisible by 4. I think we're in the same boat, though. We just can't come up with a way to express what's in our heads. Whenever I'm in a situation like this (which happens pretty often), I sit back and watch the rest of you all solve where I have struggled. Godspeed, my friends! Should anything pop into my head, I will be certain to post it. EDIT: What if we rename the plates to a, a+x, and a+y (or a+x+y)?
  23. Digging Probabilities

    So the other day I was watching a speedrun of the Legend of Zelda: Ocarina of Time (a speedrun is a playthrough of a game with the intent of beating the game as fast as possible). In one particular part of the game, the player is forced to follow around a gravedigger as he digs up various holes. There is one particular outcome that is desired (the "jackpot" of the game, essentially): a permanent health upgrade. There are also three undesirable outcomes that only give out money: a green rupee (the least valuable prize, pretty much $1), a blue rupee (a fairly desirable prize, say about $5), and a red rupee (a very desirable prize, say $20). Here are the rules: The chances of digging up a green rupee is 40%, a blue rupee 30%, a red rupee 20%, and the health upgrade 10%. If the gravedigger has dug up eight green rupees already, and he would dig up a ninth green rupee this time around, he will instead dig up the health upgrade. If the gravedigger has dug up four blue rupees already and he would dig up a fifth this time around, he will instead dig up a green rupee. If the gravedigger has dug up two red rupees already and he would dig up a third this time around, he will instead dig up a blue rupee. As a result, the maximum number of attempts to dig up the health upgrade is fifteen. Example: four blue rupees and two red rupees have been dug up. If the gravedigger hit the 20% chance to dig up a red rupee on his seventh total attempt, he would instead dig up a blue rupee. However, four blue rupees have already been dug up, so he would actually dig up a green rupee. After digging up the health upgrade, the game is over. Three questions: One: what is the probability that it will take a player the maximum number of tries to dig up the most desirable outcome (the health upgrade)? Two: What is the expected average number of tries for the health upgrade? Three: There exists two methods to play this minigame. First is the method described above. Second is that after the first dig, the player exits and reenters the area. This resets the green/blue/red rupee counter, but also allows the player to try another dig far faster. This introduces the potential for an infinite number of tries (the world record currently sits at about 100, which is pretty unlucky to say the least). Say it takes a player ten seconds between digs using the first method, and five seconds between digs using the second method. On average, which one will get you the health upgrade the fastest? Disclaimer: I've got an answer to the first question and maybe the second, but I've got no clue for the third.
  24. Jelly beans join the clean plate club

    Especially when you consider as well that at any point you've got an extra plate either to transfer to or to transfer from (although that does complicate things a bit, it also ensures that pairs such as {4,8} are solvable).
  25. Jelly beans join the clean plate club

    Here’s a tiny observation about what the next to last step looks like.
  26. Jelly beans join the clean plate club

    But no matter what you'll always have at least one pair whose average is even. Try it: pick any three numbers such that the average of any two is odd. I'm fairly sure it's impossible, since between a+b, b+c, and a+c at least one is guaranteed to be even. Do you have another counterexample of two numbers whose average is even but can't be reached via doubling?
  27. Last week
  28. Waiting, again II

    much more clearly stated than my babbling. I’m tickled that I’ve shown that it can be evaluated one flip at a time, based merely on the parity of contiguous Hs. here is my argument, expressed by plagiarizing your expression: Let e be the expected number of flips from the initial (even) state There are two states that are easy to analyze and cover all the possibilities: E is the initial state, it also represents the state of having seen an even number of H (including zero), since the beginning or the most recent T. O is the odd state, representing the fact of having seen an odd contiguous run of H. State E requires e more flips. In this state, H changes to state O, while T remains in state E State O requires o more flips. In this state, H changes to state E, while T terminates with a win. That allows us to write an expression for x as the sum of these terms, weighted by their respective probabilities, all 1/2. e = 1/2{1+e} + 1/2{1+o} o = 1/2{1+e} + 1/2 substitute o into e e = 1/2{1+e} + 1/2{1+1/2{1+e} + 1/2} = 1/2{1+e} + 1/2+1/4+e/4 + 1/4 = 3/2 + 3e/4 e/4 = 3/2 e = 6
  29. Born on a Wednesday

    (1) (2)
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